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\centerline{MATH 242 -- Principles of Analysis}
\centerline{Solutions for Problem Set 1 -- due:  Feb. 1}
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\noindent
{\it `A` Section}
\bigskip
\item{1.}  Let $A = \{x \in {\bf R} \mid x^2 - 5x + 6 = 0\}$, $B = (0,4) = \{x \in {\bf R} \mid 0 < x < 4\}$ and $C = \{{x\over x^2 + 1} \mid x \in {\bf R}\}$  (Note:  $C$ is the range of the function $f$ defined by  $f(x) = {x\over x^2 + 1}$.)
\itemitem{a.}  Express the set $C$ as a union of one or more closed intervals $[a,b]$ in ${\bf R}$. (Note:  You should use facts from calculus to solve this.  Don't worry that we have not justified them yet.)
\itemitem{} {\it Solution:}  The function $f(x) = {x\over x^2 + 1}$ has $f'(x) = {1 - x^2\over x^2 + 1}$.
This is $ = 0$ at $x = \pm 1$.  Moreover $f'(x) < 0$ for $x < -1$, $f'(x) > 0$ for $-1 < x < 1$ and
$f'(x) < 0$ for $x > 1$.  Therefore, at $x = -1$, $f$ has a local minimum with $f(-1) = -1/2$.  Similarly, 
at $x = 1$, $f$ has a local maximum with $f(1) = 1/2$.  We also see $\lim_{x \to \pm \infty} f(x) = 0$.
Hence $f(-1) = -1/2$ is also an absolute minimum, and $f(1) = 1/2$ is also an absolute maximum.  We will show
later in the course that every $y$ with $-1/2 < y < 1/2$ is also in the range.  Hence $C = [-1/2,1/2]$. 
\itemitem{b.}  Find the sets $A \cap C$ and $B\cap C$.
\itemitem{} {\it Solution:}  Since $A = \{2,3\}$, we see that $A \cap C = \emptyset$ and
$B \cap C = (0,1/2]$.
\itemitem{c.}  Find the sets $B\cup A$ and $B\cup C$ and express as unions of intervals in ${\bf R}$.
\itemitem{} {\it Solution:}  We have $B \cup A = (0,4) = B$, since $A \subset B$.  Then by part a,
$B \cup C = (0,4) \cup [-1/2,1/2] = [-1/2,4)$.    
\item{2.}    Let $B_n = \{1,1/4,1/9,\ldots,1/n^2\}$ for each natural number $n \ge 1$. 
What are $\cap_{n=1}^\infty B_n$ and $\cup_{n=1}^\infty B_n$?
\item{} {\it Solution:}  The union, $\cup_{n=1}^{\infty} B_n$, is the set
$$\{1/n^2 \mid n \ge 1\}.$$
The intersection, $\cap_{n=1}^{\infty} B_n$, is the set $\{1\}$, since that is the 
only element in $B_n$ for all $n \ge 1$.    
\item{3.}  Let $I_n = [-1/n,1/n]$ for any $n \ge 1$.
 What are $\cap_{n=1}^\infty I_n$ and $\cup_{n=1}^\infty I_n$.
(Explain your reasoning intuitively.)
\item{} {\it Solution:}  Note first that $I_m \subset I_n$ whenever $m > n$.  This shows that the union
is the same as $I_1 = [-1,1]$.  The intersection contains only $0$.  We will see in about
a week how to justify the claim that for any real $a > 0$, there is some $n \ge 1$ such 
that $1/n < a$.  Hence $a$ is not in the intersection.  The same is true on the negative
side:  for any $b < 0$, there exists some $n \ge 1$ such that $b < -1/n$.  Hence $b$ is 
not in the intersection either.  This leaves only $0$ which does satisfy $-1/n < 0 < 1/n$
for all $n \ge 1$.
\item{4.}  Let $f : {\bf R} \to {\bf R}$ be the function defined by $f(x) = x^2 - 4x + 1$.
\itemitem{a.}  Is $f$ one-to-one?  Why or why not?
\itemitem{} {\it Solution:}  By completing the square, we see $x^2 - 4x + 1 = (x - 2)^2 - 3$.
From this we can see for instance that $f(3) = -2 = f(1)$.  Therefore $f$ is not one-to-one.
We also see that the graph $y = f(x)$ is a shifted parabola with vertex at $(2,-3)$.  This
fact can be used to see parts of what we are saying in the later parts of the problem.  
\itemitem{b.}  Is $f$ onto?  Why or why not?
\itemitem{} {\it Solution:}  By the same computations as for part a, we see that 
$f(x) \ge -3$ for all $x$.  Therefore $f$ is not onto ${\bf R}$.  
\itemitem{c.}  If $I = (1,3)$, what is the set $f(I)$?  Explain.
\itemitem{} {\it Solution:}  $f$ has a local and global minimum at $f(2) = -3$.  Hence
$f((1,3)) = [-3,-2)$.  
\itemitem{d.}  If $J = (5,6)$, what is the set $f^{-1}(J)$.  Explain.
\itemitem{} {\it Solution:}  We have $f(x) = x^2 - 4x + 1 = (x - 2)^2 - 3 = 5$ when $x = 2 \pm \sqrt{8}$.
Similarly, $f(x) = 6$ when $x = 2\pm 3 = -1,5$.  Hence $f^{-1}(J)$ is the 
 union of the two intervals $f^{-1}(J) = (-1, 2 - \sqrt{8}) \cup (2+\sqrt{8},5)$.
\bigskip
\noindent
{\it `B' Section}
\bigskip
\item{1.}  Prove part (f) of Theorem 1.1.3 in the text.  These are the {\it De Morgan Laws} for
complements.
\item{} {\it Solution:}  We show $(A \cap B)^c = A^c \cup B^c$.  Let $x \in (A \cap B)^c$, then
$x \notin A\cap B$, which says $x \notin A$ {\it or} $x \notin B$.  But then $x \in A^c \cup B^c$, 
and it follows that $(A \cap B)^c \subset A^c \cup B^c$.
Conversely, if $x \in A^c \cup B^c$, then $x \notin A$ or $x \notin B$.  This shows $x \notin A\cap B$, 
so $x \in (A\cap B)^c$, and it follows that $A^c \cup B^c \subset (A \cap B)^c$.  Since we have both 
inclusions, $(A \cap B)^c = A^c \cup B^c$.  The second statement $(A \cup B)^c = A^c \cap B^c$ is 
proved similarly.  
\item{2.}  Let $A$ and $B$ be arbitrary sets.  Does $B = A - (A - B)$, as we might expect
if we looked at the formula through the lens of ordinary algebra?  If this is always true,
prove it; if it is not, give both a counterexample (an example where the formula is not true),
and a correct statement with proof.
\item{} {\it Solution:}  This is {\it not true} in general as the following counterexample
shows.  Let $A = \{a\}$ and let $B = \{b\}$ (with $a \ne b$).  Then $A - B = \{a\} = A$, so 
$A - (A - B) = \emptyset \ne B$.  The statement that is true here is that $A - (A - B) = A\cap B$.  
To prove this quickly, the best way is probably to use the De Morgan Laws from question 1 and
other parts of Theorem 1.1.3 in the text.
We have 
$$\eqalign{A - (A - B) &= A \cap (A - B)^c\cr
&= A \cap (A \cap B^c)^c\cr
&= A \cap (A^c \cup (B^c)^c)\qquad {\rm (by\ 1.1.3\ (f))}\cr
&= A \cap (A^c \cup B)\qquad {\rm (by\ 1.1.3\ (a))}\cr 
&= (A \cap A^c) \cup (A \cap B)\qquad {\rm (by\ 1.1.3\ (e))}\cr
&= \emptyset \cup (A \cap B)\cr
&= A \cap B.\cr}$$
\item{3.}  Let $f : A \to B$ be a function.
\itemitem{a.} Let $C, D$ be subsets of $A$.  Is it always
true that $f(C\cup D) = f(C) \cup f(D)$?  If this is always true prove it; if it is not,
give a counterexample.
\itemitem{} {\it Solution:}  This statement is always true.  We can prove it as follows.
If $x \in C \cup D$, then $x \in C$ or $x \in D$.  Hence $f(x) \in f(C)$ or $f(x) \in f(D)$.
It follows that $f(x) \in f(C) \cup f(D)$, so $f(C\cup D) \subset f(C) \cup f(D)$.  Conversely, if $y \in f(C) \cup f(D)$, then
$y \in f(C)$ or $y \in f(D)$.  So $y = f(x)$ for some $x \in C$ or $y = f(x)$ for some $x \in D$.
It follows that $y \in f(C \cup D)$, so $f(C) \cup f(D) \subset f(C\cup D)$.  This shows 
the equality.
\itemitem{b.}  Show that $f$ is onto if and only if $f(f^{-1}(E)) = E$ for all subsets $E$ of $B$.
\itemitem{} {\it Solution:}  Suppose that $f(f^{-1}(E)) = E$ for all subsets $E \subset B$.
Let $b \in B$ and $E = \{b\}$, then $f^{-1}(E) \ne \emptyset$ since $f(f^{-1}(E)) = E$. 
Thus there is some $a \in f^{-1}(E)$, so $f(a) = b$.  Since this is true for all $b \in B$, 
$f$ is onto.  Conversely, if $f$ is onto, we must show $f(f^{-1}(E)) = E$ for all subsets
$E \subset B$.  So let $E$ be an arbitrary subset of $B$. The definition of the inverse image says $f(f^{-1}(E)) \subset E$ for
all mappings $f$ (that is, even without the assumption that $f$ is onto).  If in addition
we know that $f$ is onto, we have that for all $b \in E$, there exist $a \in A$ such
that $f(a) = b$, and hence that those $a \in f^{-1}(E)$.  It follows that $E \subset f(f^{-1}(E))$
when $f$ is onto.  Hence if $f$ is onto, then $f(f^{-1}(E)) = E$.  
\item{4.}  Let $f : A \to B$ and $g : B \to C$.
\itemitem{a.}  Show that if $f$ and $g$ are both one-to-one, then $g \circ f : A \to C$ is
also one-to-one.
\itemitem{} {\it Solution:}  Let $(g\circ f)(x) = (g\circ f)(y)$ for some $x,y \in A$.  
Then $g(f(x)) = g(f(y))$.  Since $g$ is assumed to be one-to-one, we have $f(x) = f(y)$.
But then, because $f$ is assumed to be one-to-one, $x = y$.  Therefore, $g\circ f$ is 
one-to-one.   
\itemitem{b.}  Is the converse of the statement in part a true?  That is, if you know
that $g \circ f$ is one-to-one, does it follow that $f$ and $g$ are one-to-one?  Prove
or find a counterexample.
\itemitem{} {\it Solution:}  This statement is {\it not true}.  For instance, consider
$f : \{b,c\} \to \{b,c\}$ defined by $f(b) = b$ and $f(c) = c$.  Also
let $g : \{a,b,c\} \to \{b,c\}$ by $g(a) = g(b) = b$ and $g(c) = c$.  Then $g\circ f : \{b,c\} \to \{b,c\}$
satisfies $(g\circ f)(b) = b$ and $(g \circ f)(c) = c$ so $g \circ f$ is one-to-one.
However, $g$ is not one-to-one.  (The statement that is true here is that $f$ must be
one-to-one and $g$ must be one-to-one when restricted to the range of $f$.)
\bye
